Lagrange Interpolation: Solved Examples & Simple Explanation

Hey guys! Let's dive into the Lagrange Interpolation Method with some solved examples. If you've ever wondered how to estimate values between known data points, you're in the right place. Lagrange interpolation is a powerful technique for constructing a polynomial that passes exactly through a given set of points. This method is incredibly useful in various fields, including numerical analysis, data fitting, and computer graphics. We'll break down the concept, walk through the formulas, and solve a couple of examples to make sure you've got a solid understanding. So, grab your coffee, and let’s get started!

What is Lagrange Interpolation?

Lagrange interpolation is a method of finding a polynomial which takes certain values at specified points. Given a set of data points $(x_0, y_0), (x_1, y_1), ..., (x_n, y_n)$, the Lagrange interpolation polynomial $L(x)$ is a polynomial of degree at most $n$ that passes through all these points. In simpler terms, imagine you have a few points plotted on a graph, and you want to draw a smooth curve (a polynomial) that connects all those points. Lagrange interpolation gives you a way to find the equation of that curve.

The formula for the Lagrange interpolation polynomial is given by:

$$L(x) = \sum_{i=0}^{n} y_i \cdot l_i(x)$$

where $l_i(x)$ are the Lagrange basis polynomials, defined as:

$$l_i(x) = \prod_{j=0, j \neq i}^{n} \frac{x - x_j}{x_i - x_j}$$

Let's break this down further. The Lagrange basis polynomials $l_i(x)$ are constructed in such a way that $l_i(x_i) = 1$ and $l_i(x_j) = 0$ for $i \neq j$. This ensures that when you evaluate $L(x)$ at any of the given points $x_i$, you get the corresponding $y_i$ value. Think of each $l_i(x)$ as a blending function that gives full weight to the $y_i$ value at $x_i$ and zero weight at all other given points.

The main advantage of Lagrange interpolation is its simplicity and ease of implementation. You don't need to solve a system of equations as you would in other interpolation methods. However, a significant drawback is that adding a new data point requires recomputing the entire polynomial, which can be computationally expensive for large datasets. Despite this, Lagrange interpolation is a valuable tool in many applications, especially when dealing with a small number of data points or when simplicity is more important than computational efficiency.

Key Concepts

Before we dive into the examples, let's recap the key concepts:

• Data Points: A set of points $(x_i, y_i)$ through which the polynomial must pass.
• Lagrange Basis Polynomials: The polynomials $l_i(x)$ that form the building blocks of the Lagrange interpolation polynomial.
• Interpolation Polynomial: The polynomial $L(x)$ that approximates the function based on the given data points.

Make sure you understand these concepts before moving on to the examples. Trust me, it will make everything much easier!

Solved Examples

Okay, let's get our hands dirty with some examples! We'll start with a simple example and then move on to a slightly more complex one. Each example will walk you through the steps of constructing the Lagrange interpolation polynomial. By the end of these examples, you'll be confident in your ability to apply this method.

Example 1: Simple Interpolation

Problem: Find the Lagrange interpolation polynomial that passes through the points $(1, 3)$, $(2, 1)$, and $(3, 4)$.

Solution:

We have three data points: $(x_0, y_0) = (1, 3)$, $(x_1, y_1) = (2, 1)$, and $(x_2, y_2) = (3, 4)$.

First, we need to find the Lagrange basis polynomials $l_0(x)$, $l_1(x)$, and $l_2(x)$.

$$l_0(x) = \frac{(x - x_1)(x - x_2)}{(x_0 - x_1)(x_0 - x_2)} = \frac{(x - 2)(x - 3)}{(1 - 2)(1 - 3)} = \frac{(x - 2)(x - 3)}{(-1)(-2)} = \frac{(x - 2)(x - 3)}{2}$$

$$l_1(x) = \frac{(x - x_0)(x - x_2)}{(x_1 - x_0)(x_1 - x_2)} = \frac{(x - 1)(x - 3)}{(2 - 1)(2 - 3)} = \frac{(x - 1)(x - 3)}{(1)(-1)} = -(x - 1)(x - 3)$$

$$l_2(x) = \frac{(x - x_0)(x - x_1)}{(x_2 - x_0)(x_2 - x_1)} = \frac{(x - 1)(x - 2)}{(3 - 1)(3 - 2)} = \frac{(x - 1)(x - 2)}{(2)(1)} = \frac{(x - 1)(x - 2)}{2}$$

Now, we can construct the Lagrange interpolation polynomial $L(x)$:

$$L(x) = y_0 \cdot l_0(x) + y_1 \cdot l_1(x) + y_2 \cdot l_2(x)$$

$$L(x) = 3 \cdot \frac{(x - 2)(x - 3)}{2} + 1 \cdot [-(x - 1)(x - 3)] + 4 \cdot \frac{(x - 1)(x - 2)}{2}$$

Now, let's simplify this expression:

$$L(x) = \frac{3}{2}(x^2 - 5x + 6) - (x^2 - 4x + 3) + 2(x^2 - 3x + 2)$$

$$L(x) = \frac{3}{2}x^2 - \frac{15}{2}x + 9 - x^2 + 4x - 3 + 2x^2 - 6x + 4$$

Combine like terms:

$$L(x) = (\frac{3}{2} - 1 + 2)x^2 + (-\frac{15}{2} + 4 - 6)x + (9 - 3 + 4)$$

$$L(x) = \frac{5}{2}x^2 - \frac{19}{2}x + 10$$

So, the Lagrange interpolation polynomial is $L(x) = \frac{5}{2}x^2 - \frac{19}{2}x + 10$.

To verify, we can plug in the original x-values and see if we get the corresponding y-values:

• $L(1) = \frac{5}{2}(1)^2 - \frac{19}{2}(1) + 10 = \frac{5}{2} - \frac{19}{2} + 10 = -7 + 10 = 3$ (Correct!)
• $L(2) = \frac{5}{2}(2)^2 - \frac{19}{2}(2) + 10 = 10 - 19 + 10 = 1$ (Correct!)
• $L(3) = \frac{5}{2}(3)^2 - \frac{19}{2}(3) + 10 = \frac{45}{2} - \frac{57}{2} + 10 = -6 + 10 = 4$ (Correct!)

Example 2: A Slightly More Complex Case

Problem: Find the Lagrange interpolation polynomial that passes through the points $(-1, 1)$, $(0, -1)$, $(2, 4)$, and $(3, 0)$.

Solution:

We have four data points: $(x_0, y_0) = (-1, 1)$, $(x_1, y_1) = (0, -1)$, $(x_2, y_2) = (2, 4)$, and $(x_3, y_3) = (3, 0)$.

Let's find the Lagrange basis polynomials $l_0(x)$, $l_1(x)$, $l_2(x)$, and $l_3(x)$.

$$l_0(x) = \frac{(x - 0)(x - 2)(x - 3)}{(-1 - 0)(-1 - 2)(-1 - 3)} = \frac{x(x - 2)(x - 3)}{(-1)(-3)(-4)} = -\frac{x(x - 2)(x - 3)}{12}$$

$$l_1(x) = \frac{(x + 1)(x - 2)(x - 3)}{(0 + 1)(0 - 2)(0 - 3)} = \frac{(x + 1)(x - 2)(x - 3)}{(1)(-2)(-3)} = \frac{(x + 1)(x - 2)(x - 3)}{6}$$

$$l_2(x) = \frac{(x + 1)(x - 0)(x - 3)}{(2 + 1)(2 - 0)(2 - 3)} = \frac{(x + 1)(x)(x - 3)}{(3)(2)(-1)} = -\frac{x(x + 1)(x - 3)}{6}$$

$$l_3(x) = \frac{(x + 1)(x - 0)(x - 2)}{(3 + 1)(3 - 0)(3 - 2)} = \frac{(x + 1)(x)(x - 2)}{(4)(3)(1)} = \frac{x(x + 1)(x - 2)}{12}$$

Now, we can construct the Lagrange interpolation polynomial $L(x)$:

$$L(x) = y_0 \cdot l_0(x) + y_1 \cdot l_1(x) + y_2 \cdot l_2(x) + y_3 \cdot l_3(x)$$

$$L(x) = 1 \cdot [-\frac{x(x - 2)(x - 3)}{12}] + (-1) \cdot [\frac{(x + 1)(x - 2)(x - 3)}{6}] + 4 \cdot [-\frac{x(x + 1)(x - 3)}{6}] + 0 \cdot [\frac{x(x + 1)(x - 2)}{12}]$$

$$L(x) = -\frac{x(x - 2)(x - 3)}{12} - \frac{(x + 1)(x - 2)(x - 3)}{6} - \frac{2x(x + 1)(x - 3)}{3}$$

Let's simplify this expression:

$$L(x) = -\frac{x(x^2 - 5x + 6)}{12} - \frac{(x + 1)(x^2 - 5x + 6)}{6} - \frac{2x(x^2 - 2x - 3)}{3}$$

$$L(x) = -\frac{x^3 - 5x^2 + 6x}{12} - \frac{x^3 - 5x^2 + 6x + x^2 - 5x + 6}{6} - \frac{2x^3 - 4x^2 - 6x}{3}$$

$$L(x) = -\frac{x^3 - 5x^2 + 6x}{12} - \frac{x^3 - 4x^2 + x + 6}{6} - \frac{2x^3 - 4x^2 - 6x}{3}$$

To combine these fractions, we need a common denominator, which is 12:

$$L(x) = \frac{-(x^3 - 5x^2 + 6x) - 2(x^3 - 4x^2 + x + 6) - 4(2x^3 - 4x^2 - 6x)}{12}$$

$$L(x) = \frac{-x^3 + 5x^2 - 6x - 2x^3 + 8x^2 - 2x - 12 - 8x^3 + 16x^2 + 24x}{12}$$

Combine like terms:

$$L(x) = \frac{(-1 - 2 - 8)x^3 + (5 + 8 + 16)x^2 + (-6 - 2 + 24)x - 12}{12}$$

$$L(x) = \frac{-11x^3 + 29x^2 + 16x - 12}{12}$$

So, the Lagrange interpolation polynomial is $L(x) = \frac{-11x^3 + 29x^2 + 16x - 12}{12}$.

Advantages and Disadvantages

Let's quickly go through some of the pros and cons of using Lagrange Interpolation.

Advantages

• Simple to Understand: The method is straightforward and easy to grasp, making it great for quick implementations.
• No Need to Solve Linear Systems: Unlike other interpolation methods, you don't need to solve complex systems of equations.
• Direct Formula: The formula is direct, meaning you can directly compute the polynomial without iterative processes.

Disadvantages

• Computational Cost: Adding new data points requires recomputing the entire polynomial, which can be expensive for large datasets.
• Not Suitable for High-Degree Polynomials: High-degree polynomials can lead to oscillations and inaccuracies, especially with equally spaced points (Runge's phenomenon).
• Lack of Error Estimation: The method doesn't provide a built-in way to estimate the error of the interpolation.

Applications of Lagrange Interpolation

Lagrange Interpolation is widely used in various fields, including:

• Numerical Analysis: For approximating functions and solving differential equations.
• Computer Graphics: For creating smooth curves and surfaces.
• Data Fitting: For fitting a polynomial to a set of data points.
• Engineering: For estimating values in experimental data.

Conclusion

So, there you have it! We've covered the basics of Lagrange Interpolation and worked through some solved examples. Hopefully, this guide has given you a solid understanding of how to use this powerful technique. Remember, while it has its limitations, Lagrange interpolation is a valuable tool for approximating functions and fitting data. Keep practicing, and you'll become a pro in no time! Happy interpolating, guys! Now you can confidently apply this method to various problems and impress your friends with your newfound knowledge.